Hey, unfortunately it’s impossible for me to draw one for you here, but you should be able to draw it yourself pretty easily!
Answer:
An atom is built with a combination of three distinct particles: electrons, protons, and neutrons. Each atom has a center nucleus, where the protons and neutrons are densely packed together. Surrounding the nucleus are a group of orbiting electrons. And here is a picture that may assist your further :
Hope this helps
Answer:
A) V_rms = 29 V
B) Vav = 0 V
Explanation:
A) We are told that;
V = V_o cos ωt
voltage amplitude; V = V_o = 41.0V
Now, the formula for the root-mean-square potential difference Vrms is given as;
V_rms = V/√2
Thus plugging in relevant values, we have;
V_rms = 41/√2
V_rms = 29 V
B) Due to the fact that the voltage is sinusoidal from the given V = V_o cos ωt, we can say that the average potential difference Vav between the two terminals of the power supply would be zero.
Thus; Vav = 0 V
Answer:
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Explanation:
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Answer:
L = 4.711 *10^{-6} kg m2/s
Explanation:


=4.5*10^-5
angular velocity

= 0.1047 rad/s
the angular momentum,



Answer:
The value is 
Explanation:
From the question we are told that
The initial pressure is
The initial temperature is ![T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K](https://tex.z-dn.net/?f=T_1%20%3D%20%2050%20%5C%20F%20%3D%20%2850%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D%20283%20%20%5C%20%20K)
The final temperature is ![T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K](https://tex.z-dn.net/?f=T_2%20%3D%20%20320%20%5C%20F%20%3D%20%28320%20-%2032%29%20%2A%20%5B%5Cfrac%7B5%7D%7B9%7D%20%5D%20%2B%20273%20%3D433%20%20%5C%20%20K)
Generally the equation for adiabatic process is mathematically represented as

=> 
Generally for a monoatomic gas 
So
![14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }](https://tex.z-dn.net/?f=14%20%2A%20283%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D%20%3DP_2%20%2A%20433%5E%7B%5Cfrac%7B%5Cfrac%7B5%7D%7B3%7D%20%7D%7B1-%20%5B%5Cfrac%7B5%7D%7B3%7D%20%5D%7D%20%7D)
=> 
=> 